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The coefficient of kinetic friction is the dimensionless quantity ${\mu }_k=|f_k|/N$ for sliding contact with positive normal force $N$ in the ideal kinetic friction model. It characterizes the pair of contacting surfaces and their condition. A constant ${\mu }_k$ means the model neglects dependence on contact slip speed; unlike the coefficient of static friction, it specifies a sliding force magnitude rather than a no-slip upper bound.

The ratio compares the tangential kinetic friction magnitude with the perpendicular normal force:Here $\mathbf{w}$ is the tangential relative velocity defining contact slip. Both numerator and denominator have the same physical dimension, so the coefficient of kinetic friction has no units. It belongs to a contact pair under specified conditions, rather than to one material alone. The coefficient of static friction instead describes the threshold $|f_s{|}_{\max }/N$ before contact slip; a pull required to start motion cannot be substituted for a sliding measurement.^[ref]

As an original model example, stipulate ${\mu }_k=0.30$ and gravitational acceleration $g=9.8m/{\mathrm{s}}^2$. An already sliding, nonrotating block of total mass $m=2\mathrm{kg}$ is pulled horizontally across a stationary horizontal surface at constant velocity. With no other vertical force, Newton's second law gives $N=mg=19.6\mathrm{N}$. Zero horizontal acceleration then requires a pulling force $F=|f_k|={\mu }_{k}mg=5.88\mathrm{N}$. Doubling the total mass to $4\mathrm{kg}$ gives $N=39.2\mathrm{N}$ and $F=11.76\mathrm{N}$. The force doubles while the coefficient of kinetic friction stays $0.30$ by assumption. These numbers are model predictions, not observations or a calibration for a particular material pair.

A constant coefficient of kinetic friction is useful because one ratio can summarize many loads, as tested by load dependence of kinetic friction. It is an empirical approximation over a stated range of normal forces, speeds and surface conditions. It implies no universal independence from speed or apparent contact area: a change of face or sliding regime can change the result. Temperature, wear and lubrication are further conditions to report. Newton's second law still applies when the constant-coefficient approximation fails.^[ref]

For fuel mass rate $s>0$ and backward relative exhaust speed $u$, a rocket sliding on a horizontal plane has remaining mass $q(t)$ and satisfies the displayed rocket equation with kinetic friction. Integration over a sliding interval gives $v(t)-v(t_0)=u\log [q(t_0)/q(t)]-\mu g(t-t_0)$. From initial rest, immediate forward sliding additionally requires thrust to overcome static friction. If both friction coefficients equal $\mu$, motion starts only when $su>\mu gq(t)$. An algebraically negative sliding speed is not the physical solution at rest.

For positive angular velocity, a motor delivering constant mechanical power $P$ produces torque $P/\omega$. With constant opposing kinetic friction torque ${\tau }_f$, angular momentum balance gives the displayed equation and a steady speed ${\omega }_{\ast }=P/{\tau }_f$. A small increase in speed reduces the driving torque below the friction torque, making this equilibrium stable within the model. The constant-power idealization is singular at exact rest and requires a separate starting regime in a real motor.

A force gauge reads the pulling force. For horizontal straight-line sliding at constant velocity, its reading equals the opposing kinetic friction magnitude, provided these are the only horizontal forces. For total supported mass $m$, including added loads, $N=mg$, so a mean sliding reading $\overline{F}$ gives ${\mu }_k=\overline{F}/(mg)$. The important readout is the sustained sliding force after the starting peak; the peak belongs to the static friction threshold.

The University of Mississippi's Experiment 6 uses a wooden block on a wooden plane, a triple-beam balance, added masses, and a dual-range 50 N force probe connected through LabPro to Logger Pro. Its procedure calibrates the probe with a $0.5\mathrm{kg}$ load on a hanger, entering the corresponding force in newtons, and measures the block's mass separately. With the plane horizontal, it starts with $0.5\mathrm{kg}$ on the block and pulls horizontally at constant speed. The recorded force trace has a starting peak; the procedure selects the subsequent constant-force interval and takes its mean. It repeats with $0.5\mathrm{kg}$ increments up to $3\mathrm{kg}$ added, then plots sliding force against normal force to estimate the coefficient of kinetic friction from the slope.^[ref]

For a replication, check the gauge's zero with the string slack and its response to known gravitational weights; zeroing alone does not check its scale. Keep the plane level, the pull parallel to it, and the block face and track region consistent. Specify the contacting materials and whether the surfaces are dry, cleaned or lubricated; repeat runs to see whether debris or wear changes the sliding force. Verify constant velocity independently from position versus time, rather than assuming that a flat force trace proves zero acceleration. These are proposed measurement controls, not additional observations from the laboratory writeup. Record the selected interval, variation between runs and gauge resolution as sources of measurement error.

For a block sliding right on a stationary horizontal plane, take rightward acceleration $a$ as positive. A gauge reading $F$ for a pull at angle $\alpha$ above horizontal has horizontal component $F\cos \alpha$ and upward component $F\sin \alpha$. With no vertical acceleration or other forces, Newton's second law givesand thereforeThis inference requires continued contact with $N>0$ and rightward contact slip, so its numerator must be nonnegative. An upward pull reduces normal force; a downward pull, represented by $\alpha <0$, increases it. The horizontal constant-velocity formula is the special case $\alpha =0$, $a=0$.

For the original $2\mathrm{kg}$, ${\mu }_k=0.30$ model, an upward pull at $\alpha =30^{\circ }$ maintaining constant velocity needsDividing this gauge reading by $mg$ would give about $0.2953$, rather than $0.30$: the gauge reads the full inclined force, and the pull has changed the normal force.

Even with a horizontal pull, an accelerating measurement needs the $ma$ correction: ${\mu }_k=(F-ma)/(mg)$. In the same model, a horizontal $7.88\mathrm{N}$ pull produces $a=1m/{\mathrm{s}}^2$ while the block slides right; subtracting $ma=2\mathrm{N}$ recovers the $5.88\mathrm{N}$ kinetic friction. Dividing the uncorrected reading by $19.6\mathrm{N}$ instead gives about $0.4020$. Measure acceleration in an inertial frame, along the sliding direction, over the same interval as the gauge reading. These corrections follow from the force balance, independently of whether the resulting coefficient of kinetic friction is constant across runs.

// Target: classical-mechanics.bigb

An experimentally useful question is whether the disappearing oscillation follows viscous damping, and whether increasing the brake changes $b$ rather than $m$ or $k$. A published teaching protocol uses a cart on an air track, attached between two springs, with ceramic magnets providing eddy-current damping and an ultrasonic motion sensor recording position against time. Replace the magnets with equal masses in the baseline trial so the moving mass stays fixed. Obtain each spring constant from the slope of applied force versus extension, then use $k=k_1+k_2$. Time many baseline cycles to check ${\omega }_0\simeq \sqrt{k/m}$. Compare one-magnet and two-magnet free decays; the protocol calls for a fit of the full trace including its offset. ^[ref]

The following additional controls make the inference testable. Level the air track and hold its air supply fixed. Check the ultrasonic motion sensor at several stationary flag positions against a ruler, including the full planned travel range; record the offset and scale correction. Check elapsed acquisition time against an independent clock, and choose a sampling rate that resolves many samples within the shortest expected oscillation period. Record the resting equilibrium before and after each run. Release the cart from a small measured displacement with minimal initial velocity, with the driving apparatus inactive. Repeat at several release sizes and magnet settings while keeping the springs and total mass unchanged. These are proposed calibration and repeatability checks, not reported measurements from the source.

For underdamping, fit the recorded position to $y(t)=y_{\mathrm{eq}}+A{e}^{-\alpha t}\cos ({\omega }_{\mathrm{d}}t+\varphi )$ by nonlinear regression. The readout to inspect is a sequence of alternating extrema about $y_{\mathrm{eq}}$ with shrinking amplitudes, and the deviations of the fitted curve from the recorded points. The model predictsCompare the right-hand side, determined independently from mass and spring constants, with the fitted left-hand side. A good fit supports a constant-coefficient single normal mode over the observed range; it does not establish a universal microscopic force law. The published protocol denotes the decay rate by $b$ and the velocity-proportional coefficient by $c$; here they are $\alpha$ and $b$, respectively.

At critical damping or overdamping, fit the corresponding nonoscillatory response including its offset and both initial conditions, rather than forcing a damped cosine onto the data. Adjusting the brake can explore the transition, but the cited protocol does not establish that its one-magnet or two-magnet settings attain critical damping. A near-critical trace over a short interval can be hard to distinguish from neighbouring damping ratios.

Important limitations are finite sensor range, missed or reflected acoustic wave pulses, measurement noise near equilibrium, nonzero release velocity, and position-dependent magnetic braking. Changing the release size tests for quadratic drag or nonlinear spring response; inspecting deviations near turning points can expose kinetic friction. Slow drift of the offset can imitate a long overdamped tail. Report uncertainties in fitted rates and independently measured $m,k$; absence of a visible crossing only bounds the behaviour within the instrument's sensitivity and recording interval. No numerical experimental results are asserted here.

Friction is a contact force tangent to the surfaces. Kinetic friction opposes contact slip; static friction maintains contact without contact slip, including rolling without slipping. The perpendicular contact component is the normal force. These two components answer different questions: how strongly the surfaces press together, and how strongly they resist sliding past one another.

The ideal Coulomb dry-contact model combines a bound on static friction with a sliding law for kinetic friction. Its parameters are the coefficient of static friction and the coefficient of kinetic friction; constant coefficients are an empirical approximation, rather than a consequence of Newton's laws of motion. Surface condition matters, and the static coefficient need not universally exceed the kinetic coefficient. ^[ref]

The model's historical setting was practical machinery: Coulomb's Théorie des machines simples investigated friction and the stiffness of ropes. The Académie's bibliography records its 1781 prize and its appearance in volume 10, pages 161–332, of the Recueil des savants étrangers. ^[ref]

In the ideal kinetic friction model, the loss of combined mechanical energy of the two contacting bodies becomes thermal internal energy at rateThis follows by adding the contact powers $f_{k}v$ and $-f_{k}U$ from Newton's third law. It depends on contact slip, whereas the contact power received by either body separately depends on its velocity in the chosen inertial frame. Thus positive frictional heating is compatible with increasing kinetic energy of one body, as the moving-belt example in kinetic friction shows. Heating need not cause a temperature rise if thermal energy escapes by heat transfer.

Choose a unit vector ${\mathbf{e}}_d$ pointing downhill and a perpendicular unit vector ${\mathbf{e}}_n$ pointing outward from the inclined plane. The gravitational weight points vertically downward. Its scalar projections onto these directions giveThe angle between the downward vertical and the inward normal is $\theta$: the inward component uses cosine, and the downhill component uses sine. These are two components of one force. They replace its vertical arrow in a component calculation; adding them to that same arrow would count gravitational weight twice.

Continued contact with a stationary straight inclined plane gives zero perpendicular acceleration. The perpendicular component of Newton's second law therefore givesThis is why the normal force generally differs from $mg$: it balances the inward component, rather than the whole gravitational weight. An extra lifting force, a curved surface, or motion of the support would require a new perpendicular balance.^[ref]

The same geometry explains the ramp's usefulness. A frictionless block pulled uphill at constant velocity needs a parallel force $P=mg\sin \theta$. Travelling a distance $L$ raises it through height $h=L\sin \theta$, so the pulling work is $PL=mgh$, its gain in gravitational potential energy. The smaller force costs a larger displacement; the work-energy theorem keeps the accounting consistent.

An inclined plane is a flat surface tilted through an angle $\theta$ above horizontal. A ramp exchanges a smaller lifting force for a longer travel distance; for a released body, it reduces the downhill component of gravitational weight and makes the resulting acceleration easier to measure. Its simplest dynamical model is a nonrotating block on a stationary, rigid surface, with constant mass $m>0$, uniform gravitational acceleration of magnitude $g>0$, and $0\le \theta <\pi /2$. Assume continued contact, negligible drag force, no tipping, and no applied pull. The only forces are gravitational weight, the normal force, and any tangential friction.

The central distinction is between a block resting on the inclined plane and one already sliding. Static friction supplies the force needed to maintain rest, up to its bound; kinetic friction has a prescribed magnitude in the ideal constant-coefficient model. These laws are empirical approximations whose range depends on the contacting surfaces.^[ref]

Kinetic friction is friction during nonzero contact slip. For one-dimensional motion along a surface, choose right as positive and write $v$ for the body's contact-point velocity and $U$ for the surface's contact-point velocity, both measured in the same inertial frame. The signed relative velocity is $w=v-U$. The ideal Coulomb law for the tangential force on the body isHere $N\ge 0$ is the magnitude of the normal force, ${\mu }_k\ge 0$ is the coefficient of kinetic friction, and $\mathrm{sgn}$ is the sign function. Thus kinetic friction opposes contact slip, which can differ from the body's direction of motion. In several tangential directions the same law reads ${\mathbf{f}}_k=-{\mu }_{k}N\mathbf{w}/|\mathbf{w}|$ for a nonzero tangential relative velocity $\mathbf{w}$. The magnitude law is documented in Ling, Sanny and Moebs, OpenStax University Physics 1, §6.2: static and kinetic friction.

For a stipulated toy model, take a nonrotating box with mass $m=1\mathrm{kg}$, normal force $N=10\mathrm{N}$, coefficient of static friction ${\mu }_s=0.5$, and coefficient of kinetic friction ${\mu }_k=0.3$. These are assumed model parameters, rather than measured properties of any particular pair of materials; the prescribed normal force is consistent with horizontal support under $g=10m/{\mathrm{s}}^2$ and no other vertical forces. On stationary ground, a box already sliding right has $f_k=-3\mathrm{N}$. With an applied horizontal force $F=6\mathrm{N}$ to the right, Newton's second law gives $a=(F+f_k)/m=3m/{\mathrm{s}}^2$, for as long as rightward contact slip persists.

Direction is clearer by comparing two instantaneous states of this same model. If the box has $v=1m/s$ on stationary ground, $U=0$ and $w=+1m/s$, so kinetic friction acts left with magnitude $3\mathrm{N}$. If the box still has $v=1m/s$ but a belt beneath it has $U=2m/s$, then $w=-1m/s$: kinetic friction acts right with magnitude $3\mathrm{N}$, even while the box moves right.

In the belt case, the power delivered to the box is $P_{\mathrm{box}}=f_{k}v=+3\mathrm{W}$. With no other horizontal force, its kinetic energy therefore increases by the work-energy theorem. By Newton's third law, the box exerts $-f_k$ on the belt, so the contact power delivered to the belt is $P_{\mathrm{belt}}=-f_{k}U=-6\mathrm{W}$. Their sum is $-3\mathrm{W}$: frictional heating produces thermal internal energy at $3\mathrm{W}$. An ideal motor maintaining the prescribed belt velocity supplies $+6\mathrm{W}$ against this contact load, neglecting other losses, so its input divides into $3\mathrm{W}$ of box kinetic energy and $3\mathrm{W}$ of frictional heating. These are instantaneous rates, since the box's velocity subsequently changes. This balance follows directly from power as force times velocity and conservation of energy; the power law is given in Ling, Sanny and Moebs, OpenStax University Physics 1, §7.4: power.

When contact slip vanishes, including a transition to rolling without slipping, the nonzero-slip law no longer applies. Determine the required static friction from the no-slip condition and the equations of motion, then test its bound. Substituting $\mathrm{sgn}(0)=0$ into the sliding law would incorrectly exclude a nonzero static friction.

Let $s=m_0/T$ be the constant fuel mass rate, and let $q(t)=M+m_0-st$ be the remaining rocket mass. Take positive motion opposite to the exhaust direction. Over a short interval $dt$, momentum conservation including the mass $sdt$ expelled at velocity $v-u$ givesConsequently, while the rocket slides forward,This is the constant-burn rocket with ground friction. Assuming forward sliding begins immediately and persists, integration from rest givesand henceA necessary physical assumption is missing from the printed request. With a common static friction and kinetic friction coefficient $\mu >0$, immediate motion requires $su\ge \mu g(M+m_0)$; equality allows motion as soon as fuel loss lowers the normal force. Then the acceleration is nonnegative initially and increases thereafter, so forward sliding is consistent. Without this assumption the displayed expression need not be the final speed and can even be negative.

For completeness, the physical rest/sliding solution with these equal coefficients is as follows. If $su\le \mu gM$, the rocket never starts before burnout and its final speed is zero. If $\mu gM<su<\mu g(M+m_0)$, static friction initially balances the thrust and release occurs atIntegrating only the sliding interval yieldsThis speed is positive because, with $y=q_s/M>1$, it equals $u[\log y-1+1/y]$, whose derivative with respect to $y$ is $(y-1)/y^2>0$. The $\mu =0$ case recovers the frictionless rocket equation. If static and kinetic coefficients differ, the release condition uses the static coefficient and the subsequent integral uses the kinetic coefficient; the prompt supplies no second coefficient.

The moment of inertia of a uniform solid cylinder is taken about its symmetry axis. Write its mass density as ${\rho }_c=m/(\pi a^{2}L)$. A coaxial shell of radius $r$ and thickness $dr$ has mass $dm=2\pi {\rho }_{c}Lrdr$. ThusA hollow cylinder would have a different moment of inertia; uniform solid density is the intended assumption.

Let $s$ measure the centre's displacement uphill from its initial position, and choose signed angular velocity $\omega$ so that uphill rolling without slipping has $\dot{s}=a\omega$. Let $F$ be the signed contact force uphill. The translation and rotation equations areThe minus sign in the torque equation follows from the chosen rolling sense: uphill contact force opposes the positive spin. Under rolling without slipping, $\ddot{s}=a\dot{\omega }$. Eliminating $F$ givesUsing $s(0)=0$ and $\dot{s}(0)=V$, the no-slip motion and maximum vertical rise areIndeed the centre turns at $t=3V/(2g\sin \alpha )$, reaching $s_{\max }=3V^2/(4g\sin \alpha )$, whose vertical component is the boxed height. Conservation of energy independently gives the same answer from initial kinetic energy $m{V}^2/2+I(V/a{)}^2/2=3m{V}^2/4$.

The available static friction is at most $\mu mg\cos \alpha$. It can supply the required uphill force exactly when $mg\sin \alpha /3\le \mu mg\cos \alpha$. ThereforeInterpret the stated friction magnitude as limiting static friction and as kinetic friction during slip, with the same coefficient. Take $0<\alpha <\pi /2$, $\mu \ge 0$. In the slipping regime the contact point moves downhill relative to the slope, so kinetic friction acts uphill. To verify this direction rather than assume it unchecked, solve with $F=\mu mg\cos \alpha$. Define $b=g(\sin \alpha -\mu \cos \alpha )>0$. The result isThe centre's speed is $|V-bt|$, with the sign of $v$ specifying ascent or descent. The contact point's relative velocity isThus the assumed uphill kinetic friction is consistent throughout both ascent and descent, including when the centre's velocity changes sign. There is no later return to rolling without slipping before reaching the bottom. This is the characteristic slipping solid cylinder on an incline regime.

The centre turns at $t_{\ast }=V/b$, givingThe formulas apply up to the positive return time $t_b=2V/b$. Substitution into the signed velocity and angular velocity givesThe return spin can have either sign and vanishes at $\tan \alpha =5\mu$. It reverses when $3\mu <\tan \alpha <5\mu$, although the contact point still slips downhill. The unchanged centre speed does not mean that no energy was dissipated: the rotational kinetic energy has decreased. Indeed integrating the frictional power $Fw<0$ giveswhich agrees with the difference of the initial and final rotational kinetic energies.

On the horizontal road the total normal reaction is $Mg$. The kinetic friction force has magnitude $\mu Mg$, so Newtonian mechanics gives $\dot{u}=-\mu g$ until rest. ThusFor the wet high-speed stage, add the four tyre drags: the total linear drag is $\lambda u$, not $\lambda u/4$. Hence $M\dot{u}=-\lambda u$ and $u(t)=U{e}^{-\lambda t/M}$. At the switching speed $u_{\ast }=U/4$, the elapsed time and distance areThe remaining stage has the same constant kinetic friction deceleration as on a dry road, but starts at $U/4$. It therefore contributes time $U/(4\mu g)$ and distance $U^2/(32\mu g)$. Adding both stages of the two-stage stopping under linear drag and dry friction givesThe exponential first stage reaches its positive switching speed in finite time; it need not reach zero under linear drag.

Let $m$ be the car's mass and let $v(t)$ be its forward speed. Initially the driven contact surfaces slip backward relative to the ground because $v<r\Omega$. Thus kinetic friction on the car acts forward. On the horizontal surface the total normal force is $mg$, so Newton's second law gives $m\dot{v}=\mu mg$. While slipping persists,The contact slip speed is $r\Omega -v(t)$. Its first zero is the onset of rolling without slipping, henceThe engine maintains the prescribed wheel speed, so wheel spin-down is not part of this calculation.

On a slope of angle $\alpha >0$, directed uphill, the normal force is $mg\cos \alpha$ and gravity contributes $-mg\sin \alpha$ along the slope. Therefore the uphill acceleration while slipping is $g(\mu \cos \alpha -\sin \alpha )$. If it is positive, the onset of rolling without slipping occurs atThe inequality follows because $\mu \cos \alpha -\sin \alpha <\mu$. If $\mu \cos \alpha \le \sin \alpha$, the car does not accelerate uphill to $r\Omega$, so there is no finite catch-up time in this sliding-friction model. The finite-time conclusion presumes positive traction sufficient to climb the slope.

The original diagram has a horizontal bottle neck, so weight does not enter the axial force balance. Gauge pressure acts on the cork's end area and supplies outward force $F_g=\pi a^2{P}_g$. The radial contact force is the contact pressure times the cylindrical contact area, $N=2\pi a(l-x)P_c$. The limiting friction is $\mu N$, so outward sliding occurs whenThe assumption of equal static and kinetic friction coefficients removes a separate force drop when slipping starts.

Initially the pressure is marginal at $x=0$, so $\pi a^2{P}_g=2\pi al\mu P_c$. During extraction the contact length decreases, while the pressure force stays constant. The net force is therefore $F(x)=\pi a^2{P}_{g}x/l$. The cork's mass is $m_c=\rho \pi a^{2}l$, giving the pressure-driven cork ejection equationFor $x(0)=0$ and $\dot{x}(0)=v_0>0$, solving the constant-coefficient equation givesThe equilibrium is unstable: as the cork moves out, friction decreases and the unbalanced force grows.

Multiplying the equation of motion by $\dot{x}$ and integrating gives ${\dot{x}}^2=v_0^2+b^2{x}^2$. The contact ends at $x=l$, soEquivalently, the net work is ${\int }_0^{l}F(x)dx=\pi a^2{P}_{g}l/2$. Exactly zero initial velocity would leave the ideal marginal configuration stationary. The exit time $b^{-1}\mathrm{arsinh}(bl/v_0)$ diverges as $v_0\to 0^{+}$, even though the limiting exit speed is finite.

The density is $500\mathrm{kg}{\mathrm{m}}^{-3}$, so $v_{\mathrm{exit}}\simeq \sqrt{400000/500}=28.3\mathrm{m}{\mathrm{s}}^{-1}$. For an ideal upward launch at this speed, conservation of mechanical energy gives height above the exitThis neglects air resistance and the small gravitational work during extraction compared with the launch energy. Actual gas pressure need not stay constant, so it is an idealized estimate.

Adding a stopper that doubles the moving mass, while preserving the same pressured area and contact friction, leaves the extraction work unchanged. The squared exit speed and the ballistic height are therefore halved:

Definitions. The torque of a force $\mathbf{F}$ about an origin is $\mathbf{\tau }=\mathbf{r}\times \mathbf{F}$, where $\mathbf{r}$ runs from that origin to the force's point of application. Its component about a fixed axis changes angular momentum about that axis. The moment of inertia about that axis iswhere $s$ is perpendicular distance to the axis. It measures resistance to angular acceleration; for fixed-axis rigid rotation, ${\tau }_{\mathrm{net}}=I\alpha$. The units are $\mathrm{N}\mathrm{m}$ for torque and $\mathrm{kg}{\mathrm{m}}^2$ for moment of inertia.

Mass and axial inertia. Slice the shaft perpendicular to its common axis. A slice of radius $r(z)$ has mass $dm=\rho \pi r(z{)}^{2}dz$ and moment of inertia $dI=\frac{1}{2}r(z{)}^{2}dm$, using the moment of inertia of a uniform disc. Thus the axial moment of inertia of a solid of revolution uses the second and fourth powers of its radius:For the tapered section, take $z$ from its narrow end toward its wide end, so $r(z)=a(1+z/l)$ for $0\le z\le l$. Its contributions areThe two cylinders contribute $4\rho \pi a^{2}l$ and $\rho \pi a^{2}l$ to mass, and $8\rho \pi a^{4}l$ and $\frac{1}{2}\rho \pi a^{4}l$ to moment of inertia. Adding the three sections givesAll slice rotation axes coincide with the shaft axis, so the moments add directly.

Frictional torque. The normal load on the narrow circular end is $Mg$. With the stipulated uniform pressure, $p_N=Mg/(\pi a^2)$. A contact annulus of radius $s$ and width $ds$ has area $dA=2\pi sds$, normal force $p_{N}dA$ and tangential kinetic friction $dF=\mu p_{N}dA$. Multiplying by its lever arm $s$ and integrating gives the uniform-pressure friction torque on a circular contact:Here $G_f$ is the positive magnitude; its torque opposes the angular velocity.

Steady speed and stopping rotations. Mechanical power in rotation is $P=\tau \omega$. At steady angular velocity, the driving and frictional torques balance, so constant-power rotation against dry friction givesUsing $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $a=0.100\mathrm{m}$,Once the motor is removed, the speed-independent kinetic friction gives constant angular acceleration $-G_f/I$. Equivalently, the initial rotational kinetic energy is dissipated as frictional work. The stopping angle under constant resisting torque is thereforeThe total angular displacement corresponds toHence the shaft completesand then another $0.16668$ of a rotation. Its stopping time is $I{\omega }_{\ast }/G_f=2.05461\mathrm{s}$. Using the rounded value $g=9.8\mathrm{m}{\mathrm{s}}^{-2}$ changes the fractional rotation slightly but leaves four complete rotations.

A cylindrical plug of radius $a$, length $l$ and density $\rho$ is pushed outward by gauge pressure $P_g$. If its remaining contact length is $l-x$ and its constant radial contact pressure is $P_c$, the resisting kinetic friction is $2\pi a(l-x)\mu P_c$. At the marginal initial condition $\pi a^2{P}_g=2\pi al\mu P_c$, the net outward force becomes $\pi a^2{P}_{g}x/l$. With mass $\rho \pi a^{2}l$,Thus the exit speed at $x=l$ tends to $\sqrt{P_g/\rho }$ as $v_0\to 0^{+}$. Exactly $v_0=0$ leaves the marginal configuration stationary, and the exit time $b^{-1}\mathrm{arsinh}(bl/v_0)$ diverges in that limit. An attached extra mass changes the acceleration but not the pressure-minus-friction work, if the contact geometry is unchanged.

First test rest. With no other tangential forces, the required uphill static friction is $f_s=mg\sin \theta$. Its maximum is ${\mu }_{s}N$, where ${\mu }_s$ is the coefficient of static friction. ThusThe tangent function is increasing on this angle range, and its inverse tangent determines the last permitted resting angle ${\theta }_c$. At equality, the ideal model still permits limiting rest; a larger angle excludes it. Below the threshold, setting $f_s={\mu }_{s}N$ would incorrectly impose the maximum static friction even when less is required. Static friction on an inclined plane and inclined-plane measurement of static friction develop this onset criterion further.

For a block already sliding downhill, uphill kinetic friction has magnitude $f_k={\mu }_{k}N$, where ${\mu }_k$ is the coefficient of kinetic friction. Taking downhill as positive, Newton's second law becomesEvery term in the force balance scales with mass; division by $m>0$ removes it. Doubling mass doubles both gravitational weight and the contact forces, while leaving this acceleration unchanged if the coefficients remain unchanged. A mass-independent result is a prediction of these assumptions, rather than a universal property of every ramp.

For an original numerical model, stipulate $m=3\mathrm{kg}$, $g=9.8m/{\mathrm{s}}^2$, ${\mu }_s=0.45$ and ${\mu }_k=0.30$. The critical angle is ${\theta }_c\simeq 24.23^{\circ }$. At $20^{\circ }$, the normal force is $27.627\mathrm{N}$; rest requires $10.055\mathrm{N}$ of uphill static friction, below its $12.432\mathrm{N}$ bound. A resting block can stay there. If it is already sliding downhill, however, the kinetic friction is only $8.288\mathrm{N}$, giving downhill acceleration $a\simeq 0.5891m/{\mathrm{s}}^2$. Specifying the initial state matters.

At $30^{\circ }$, the downhill component of gravitational weight is $14.700\mathrm{N}$, while $N\simeq 25.461\mathrm{N}$ and the static friction bound is only $11.458\mathrm{N}$. Rest is impossible. During downhill sliding,Released from rest on a sufficiently long ramp, this ideal block covers its first $1.00\mathrm{m}$ in $t=\sqrt{2L/a}\simeq 0.9218\mathrm{s}$, reaching speed $v=\sqrt{2aL}\simeq 2.1697m/s$. As an independent energy calculation, the lost gravitational potential energy is $mgL\sin \theta =14.700\mathrm{J}$ and kinetic friction removes $f_{k}L\simeq 7.638\mathrm{J}$ of mechanical energy. The remaining $7.062\mathrm{J}$ equals $m{v}^2/2$ by the work-energy theorem.

At $\theta =\arctan {\mu }_k$, downhill sliding has zero acceleration and therefore constant velocity. Below that angle, a downhill-moving block slows; when it stops, test the static friction bound again. For uphill sliding, kinetic friction points downhill, so the signed downhill acceleration instead is $g(\sin \theta +{\mu }_k\cos \theta )$. The sign follows contact slip, and the downhill formula cannot be used unchanged after a reversal.

For a uniform solid cylinder, take $s$ positive uphill and $\omega$ positive for uphill rolling without slipping. The velocity of the contact point is $w=\dot{s}-a\omega$. Uphill kinetic friction gives $\ddot{s}=-g(\sin \alpha -\mu \cos \alpha )$ and $a\dot{\omega }=-2\mu g\cos \alpha$. Starting from $\dot{s}(0)=a\omega (0)$, one obtains $w=-gt(\sin \alpha -3\mu \cos \alpha )$. Thus $\tan \alpha >3\mu$ makes the assumed uphill kinetic friction self-consistent for the entire ascent and descent. Without slip the required uphill static friction is $mg\sin \alpha /3$; it is available exactly when $\tan \alpha \le 3\mu$ under the same limiting coefficient.